# What is the standard form of the equation of the parabola with a directrix at x=9 and a focus at (8,4)?

Aug 26, 2017

The standard form is: $x = - \frac{1}{2} {y}^{2} + 4 y + \frac{1}{2}$

#### Explanation:

Because the directrix is a vertical line, one knows that the vertex form of the equation for the parabola is:

$x = \frac{1}{4 f} {\left(y - k\right)}^{2} + h \text{ }$

where $\left(h , k\right)$ is the vertex and $f$ is the signed horizontal distance from the vertex to the focus.

The x coordinate of the vertex halfway between the directrix and the focus :

$h = \frac{9 + 8}{2}$

$h = \frac{17}{2}$

Substitute into equation :

$x = \frac{1}{4 f} {\left(y - k\right)}^{2} + \frac{17}{2} \text{ }$

The y coordinate of the vertex is the same as the y coordinate of the focus:

$k = 4$

Substitute into equation :

$x = \frac{1}{4 f} {\left(y - 4\right)}^{2} + \frac{17}{2} \text{ }$

The value of $f$ is the signed horizontal distance from the vertex to the focus#

$f = 8 - \frac{17}{2}$

$f = - \frac{1}{2}$

Substitute into equation :

$x = \frac{1}{4 \left(- \frac{1}{2}\right)} {\left(y - 4\right)}^{2} + \frac{17}{2}$

This is the vertex form:

$x = - \frac{1}{2} {\left(y - 4\right)}^{2} + \frac{17}{2}$

Expand the square:

$x = - \frac{1}{2} \left({y}^{2} - 8 y + 16\right) + \frac{17}{2}$

Use the distributive property:

$x = - \frac{1}{2} {y}^{2} + 4 y - 8 + \frac{17}{2}$

Combine like terms:

$x = - \frac{1}{2} {y}^{2} + 4 y + \frac{1}{2}$

Here is a graph of the standard form, the focus, the vertex, and the directrix: 